How Do You Find Alternate Polar Coordinates with Different Signs for R?

[stunner5000pt]

Homework Statement

Find two other pairs of polar coordinates of the given polar coordinate, one with r > 0 and one with r < 0. Then plot the point.

(2, 5π/3)

Homework Equations

I don't there are any.

The Attempt at a Solution

I'm not completely sure of how to do this actually.

I know that we can add 2 pi to the answer to get the coordinates for r > 0 this gives an answer of (2,\frac{11\pi}{3}

But I'm not sure about the r<0 option. Would I simply have to subtract 2 pi from 5 pi/3? In addition, would the r value change to the negative sign?

Thanks for your help.

 

[tiny-tim]

hi stunner5000pt!

Find two other pairs of polar coordinates of the given polar coordinate, one with r > 0and one with r < 0.Then plot the point.

(2, 5π/3)

are you sure that isn't θ > 0 and θ < 0 ?

 

[stunner5000pt]

hi stunner5000pt!


are you sure that isn't θ > 0 and θ < 0 ?

I chekced again and yes it is R and not theta ...

 

[eumyang]

are you sure that isn't θ > 0 and θ < 0 ?

But I'm not sure about the r <0 option. Would I simply have to subtract 2 pi from 5 pi/3? In addition, would the r value change to the negative sign?

r can be negative. Think of yourself being in the origin of the coordinate plane, facing the positive y direction. If you walk in the positive y direction 5 units, then where you end up will correspond to
\left( 5, \frac{\pi}{2} \right)
in polar coordinates.

Think of yourself back in the origin, facing the positive y direction. Now walk backwards 5 units. By "walking backwards" your r is negative. Since you are still facing the positive y direction, this point will correspond to
\left( -5, \frac{\pi}{2} \right)
in polar coordinates. But this particular point is the same as facing the negative y direction and walking forwards 5 units, or
\left( 5, \frac{3\pi}{2} \right)

Now see if you can answer your question.

 

[stunner5000pt]

r can be negative. Think of yourself being in the origin of the coordinate plane, facing the positive y direction. If you walk in the positive y direction 5 units, then where you end up will correspond to
\left( 5, \frac{\pi}{2} \right)
in polar coordinates.

Think of yourself back in the origin, facing the positive y direction. Now walk backwards 5 units. By "walking backwards" your r is negative. Since you are still facing the positive y direction, this point will correspond to
\left( -5, \frac{\pi}{2} \right)
in polar coordinates. But this particular point is the same as facing the negative y direction and walking forwards 5 units, or
\left( 5, \frac{3\pi}{2} \right)

Now see if you can answer your question.

Thanks for your advice

based on what you have said, in the case of (2, 5pi/3), the moment arm for the 'backward' motion is in the second quadrant

If we 'walked' backward that would be (-2, \frac{5\pi}{3})
for the 'positive' motion it would be (2, \frac{2\pi}{3})

How does this sound?

 

[eumyang]

Thanks for your advice

based on what you have said, in the case of (2, 5pi/3), the moment arm for the 'backward' motion is in the second quadrant

If we 'walked' backward that would be (-2, \frac{5\pi}{3})
for the 'positive' motion it would be (2, \frac{2\pi}{3})

How does this sound?

Not quite. The three points you state are not the same. The three points I used in my example were not the same either - the last two formed a separate example from the first.

The question is this: which direction would you have to "face", if, when walking "backwards" 2 units, you end up at the same point as (2, 5π/3)? In other words,
\left( 2, \frac{5\pi}{3} \right) = \left( -2, ? \right)
Then you also need to consider coterminal angles.

 

[stunner5000pt]

Not quite. The three points you state are not the same. The three points I used in my example were not the same either - the last two formed a separate example from the first.

The question is this: which direction would you have to "face", if, when walking "backwards" 2 units, you end up at the same point as (2, 5π/3)? In other words,
\left( 2, \frac{5\pi}{3} \right) = \left( -2, ? \right)
Then you also need to consider coterminal angles.

if we walked backward, then the angle would be pi/3

As for the answer with r > 0, the answer wuld be (2, -pi/3)

Is that correct?

 

[SammyS]

if we walked backward, then the angle would be pi/3 not correct

As for the answer with r > 0, the answer would be (2, -pi/3) correct

Is that correct?

See the red comments.

 

[eumyang]

if we walked backward, then the angle would be pi/3

Sorry, that's still not correct. It may help if you looked at a polar coordinate grid like the one on this site.

 

[stunner5000pt]

Sorry, that's still not correct. It may help if you looked at a polar coordinate grid like the one on this site.

Hmm... ok (2, -pi/3) is correct. This is because if we have a positive moment arm, the angle would be measured negative as we rotate clockwise

for the negative angle, wouldn't the angle be the same? But since the r is negative, would that imply that if we rotate clockwise, we get a positive angle?
would that mean that for r = -2, the angle should be + pi/3?

 

[eumyang]

Hmm... ok (2, -pi/3) is correct. This is because if we have a positive moment arm, the angle would be measured negative as we rotate clockwise

for the negative angle, wouldn't the angle be the same? But since the r is negative, would that imply that if we rotate clockwise, we get a positive angle?
would that mean that for r = -2, the angle should be + pi/3?

No. See the attached image. Point A is the point you want, \left( 2, \frac{5\pi}{3} \right). Point B is \left( -2, \frac{\pi}{3} \right), the point that you are saying that is the same as point A. They are not the same.

 

[Chestermiller]

In polar coordinates, the radial coordinate r is always positive.

 

[stunner5000pt]

I am so confused

SO let's start from the top

if we consider a point \left( 2,\frac{5\pi}{3} \right)
then for r> 0 \left( 2,\frac{11 \pi}{3} \right)
and for r < 0 \left( -2,\frac{2\pi}{3} \right)

obtained r > 0 by simply rotating once (+2 pi)
obtained r < 0 by rotating half circle (- pi)

 

[skiller]

In polar coordinates, the radial coordinate r is always positive.

Who says?

I am so confused

SO let's start from the top

if we consider a point \left( 2,\frac{5\pi}{3} \right)
then for r> 0 \left( 2,\frac{11 \pi}{3} \right)
and for r < 0 \left( -2,\frac{2\pi}{3} \right)

obtained r > 0 by simply rotating once (+2 pi)
obtained r < 0 by rotating half circle (- pi)

Bingo!

 

[Chestermiller]

Who says?

Every math book that I've ever seen. I challenge you to site one single reference in which r in polar coordinates is considered anything but positive (or zero).

 

[skiller]

Every math book that I've ever seen. I challenge you to site one single reference in which r in polar coordinates is considered anything but positive (or zero).

Well apart from at least one other person on this thread saying so, and at least one link on this thread to a website saying so...

Even the Wiki page for Polar Coordinates says so. (Although, I will agree that Wiki is not always Gospel.)

 

[LCKurtz]

In polar coordinates, the radial coordinate r is always positive.

Who says?

Bingo!

Every math book that I've ever seen. I challenge you to site one single reference in which r in polar coordinates is considered anything but positive (or zero).

Plot $r = \sin (3\theta)$ for $\theta$ from $0$ to $\pi$ and you will see an nice 3 leaved rose with one leaf plotted with $r<0$.

 

[SammyS]

I am so confused
...

I don't blame you.

It's clear that this argument concerning polar coordinates is not helping you understand how to solve this problem you have been given.



Try looking at the case of r = -2 as follows.

In Cartesian coordinates, the point you're working with is \displaystyle \ (x,\, y)=\left(1,\,-\sqrt{3}\right)\ .

In general, x = r cos(θ) and y = r sin(θ).

For this point, you then have \displaystyle \ 1=(-2)\cos(\theta)\ \text{ and }\ -\sqrt{3}=(-2)\sin(\theta) \ .

This gives you \displaystyle \ \cos(\theta)=-\,\frac{1}{2}\ \text{ and }\ \sin(\theta) =\frac{\sqrt{3}}{2} \ .

Thus the angle, θ, is in the second quadrant.

 

[eumyang]

Every math book that I've ever seen. I challenge you to site one single reference in which r in polar coordinates is considered anything but positive (or zero).

Precalculus by Larson (8th ed.) refers to r as a "directed distance from O to P" (p. 777) and it states that "another way to obtain multiple representations of a point is to use negative values for r" (p. 778).

Precalculus: Graphical, Numerical, Algebraic by Demana/Waits/Foley/Kennedy (7th ed.) states that "r is the directed distance from O to P" and that "if r < 0 then P is on the terminal side of θ + π" (both on p. 534)

That's two references, and I think it's safe to say that you are mistaken.

 

[Chestermiller]

Well apart from at least one other person on this thread saying so, and at least one link on this thread to a website saying so...

Even the Wiki page for Polar Coordinates says so. (Although, I will agree that Wiki is not always Gospel.)

Actually, Wiki is a little schizophrenic on this. First the say "The radial distance ρ is the Euclidean distance from the z axis to the point P." This can only be positive. But then they say something about allowing negative values of ρ to locate a point.

In a 40 year career as an engineer/mathematician, I have never seen negative values of ρ being used in practice. So maybe negative values of ρ are only found in Ivory Tower land.

 

[stunner5000pt]

thank you all for your help.

As for r being negative... please contact the author of Stewart's Calculus (whose name is Stewart :D) and ask him what he thinks because this question was derived from that textbook and put on the online assignment for a Calculus II class